For what values of $a$ and $b$ does the series $\sum_{n\geq1}\ln^{a}(n)n^{b}$ converge?

Question

I want to know for what values of $a,b$, the series $\displaystyle \sum_{n=1}^\infty \ln^a(n) n^b$ converges. I have some specific cases, but I couldn't determinate anything in a general way. Thanks for any help.

Answer 1

Considering $$\sum_{n=2}^\infty n^b \ln^a n $$ it's clear that the sum diverges for $b>-1$ and converges for $b < -1$ irrespective of $a$. In the case $b=-1$ use Cauchy's condensation test:

$$\sum_{n=2}^\infty n^{-1} \ln^a n \ \ \text{ converges} \quad\iff\quad \sum_{n=2}^\infty 2^n\cdot 2^{-n} \ln^a (2^n) \ \ \text{ converges}$$

The sum on the right is $$\sum_{n=2}^\infty 2^n\cdot 2^{-n} \ln^a (2^n) = \ln^a (2)\cdot \sum_{n=2}^\infty n^a $$ and converges iff $a<-1$. What's left for you is to show that the preconditions for that test are met, and to work out some details of $b\neq-1$.

Taking all together, the series converges iff $b < -1 \lor (b = -1 \land a < -1)$.